5
$\begingroup$

I have been given this recently in PDE class involving the solutions to the Bessel fucntion in Sturm-Liouville form, asking for Eigenvalues and Eigenfunctions:

$ (xy')'+\lambda x y = 0 \space \space \space \space 0<a<x<b $

$ y(a)=y(b)=0 $

Here is where my problems start: If I expand the equation I get:

$ xy''+y'+\lambda x y = 0 \to x^2y''+xy'+\lambda x^2 y = 0 $ now depending on whether or not the eigenvalues are positive or negative (checking zero is not one is fairly simple) we obtain the general solution to the Bessel equation or modified Bessel equation of order zero:

$ y(x) = AJ_0(\sqrt{\lambda}x) + BY_0(\sqrt{\lambda}x) $ or

$ y(x) = AI_0(\sqrt{\lambda}x) + BK_0(\sqrt{\lambda}x) $

So there is not the matter of boundedness at zero resulting in eliminating the singular solution from the linear combination so my problem here is basically applying the boundary conditions like this:

$ AJ_0(\sqrt{\lambda}a) + BY_0(\sqrt{\lambda}a) = AJ_0(\sqrt{\lambda}b) + BY_0(\sqrt{\lambda}b) = 0 $

so this equation and deriving $ \lambda $ and the actual eigenfunctions from this is beyond my ability, and that's where I am stuck and need the help. Thanks all.

$\endgroup$
  • 1
    $\begingroup$ Can you make that last paragraph a bit clearer and more precise? I can't disentangle what the problem you're having actually is. $\endgroup$ – Chappers Oct 4 '15 at 15:19
  • $\begingroup$ @Chappers : I hope this looks more decent $\endgroup$ – kroner Oct 4 '15 at 15:28
  • 1
    $\begingroup$ AFAIK, there is no explicit form for the solution of those equations. You can get the implicit equation for the eigenvalues by ensuring that the determinant of the matrix of those two equations in $A,B$ vanishes, but no further. $\endgroup$ – Chappers Oct 4 '15 at 15:33
  • 1
    $\begingroup$ This is the relationship used for orthonormalising Bessel functions on finite intervals. $\endgroup$ – Chappers Oct 4 '15 at 15:45
  • 1
    $\begingroup$ The asymptotic behaviour is sinusoidal for both $J_0$ and $Y_0$. Hence large eigenvalue ($lambda>>0$) should resemble the particle in a box spectrum. In other cases, may be numerical methods would have to be applied. $\endgroup$ – vnd Oct 4 '15 at 15:49
6
$\begingroup$

You can show that the eigenvalue must be positive even without solving the differential equation. The Bessel equation in Sturm-Lioville form is

$${\left( {xy'} \right)^\prime } = - \lambda xy$$

Multiply by $y$ and integrate from $a$ to $b$ to obtain

$$\int_a^b {y{{\left( {xy'} \right)}^\prime }dx} = - \lambda \int_a^b {x{y^2}dx} $$

now use integration by parts for the left side

$$\left. {xyy'} \right|_a^b - \int_a^b {x{{\left( {y'} \right)}^2}dx} = - \lambda \int_a^b {x{y^2}dx} $$

using the boundary conditions and solving for $\lambda $ leads to

$$\lambda = {{\int_a^b {x{{\left( {y'} \right)}^2}dx} } \over {\int_a^b {x{y^2}dx} }}$$

considering $b>a>0$ (this must happen to satisfy the Sturm-Liouville assumption $x>0$) you can easily conclude that

$$\lambda \ge 0$$

but $\lambda $ cannot be zero too! The reason is that by the above formula $\lambda $ is zero if and only if ${y'}$ is zero and this implies $y = Const$ but by the boundary conditions you conclude that $y=0$. It means that $\lambda $ is zero if and only if $y$ is zero. Hence, not leading to a nonzero solution. Finally, we proved that

$$\lambda > 0$$

$\endgroup$
  • 1
    $\begingroup$ Apparently you can only have one marked answer simultaneously $\endgroup$ – kroner Oct 4 '15 at 17:56
5
$\begingroup$

It's not that much hard. When applying BC's you are left with

$$\left\{ \matrix{ {J_0}(\sqrt \lambda a)\,A + \,{Y_0}(\sqrt \lambda a)\,B = 0 \hfill \cr {J_0}(\sqrt \lambda b)\,A + \,{Y_0}(\sqrt \lambda b)\,B = 0 \hfill \cr} \right.$$

We want this linear system to have nonzero solutions and hence it's determinant must be zero. So we must have

$${J_0}(\sqrt \lambda a){Y_0}(\sqrt \lambda b) - {J_0}(\sqrt \lambda b){Y_0}(\sqrt \lambda a) = 0\,$$

This is the equation that we should compute eigenvalues from it. Note that this equation has infinitely many positive roots. It remains to determine the eigenfunctions. Considering the above equations we have

$$\left\{ \matrix{ A = \mu B \hfill \cr \mu = - {{{Y_0}(\sqrt \lambda a)\,} \over {{J_0}(\sqrt \lambda a)}} = - {{{Y_0}(\sqrt \lambda b)} \over {{J_0}(\sqrt \lambda b)}} \hfill \cr} \right.$$

and hence we have

$${y_\lambda }(x) = A\left( {{\mu _\lambda }{J_0}(\sqrt \lambda x) + {Y_0}(\sqrt \lambda x)} \right)$$

in order to have a beautiful final answer choose $A = -{J_0}(\sqrt \lambda a)$ and hence our final result is

$${y_\lambda }(x) = {Y_0}(\sqrt \lambda a){J_0}(\sqrt \lambda x) - {J_0}(\sqrt \lambda a){Y_0}(\sqrt \lambda x)$$

you can easily see that these eigenfunctions have zero value at $a$ and $b$.

$\endgroup$
  • $\begingroup$ Thank you for this well written answer but is there some way to tell if eigenvalues can be positive or negative? I.E to determine if it's Bessel Functions or modified Bessel functions or both? $\endgroup$ – kroner Oct 4 '15 at 15:51
  • 1
    $\begingroup$ Yeah it does. I will right it down for you as another answer. :) $\endgroup$ – H. R. Oct 4 '15 at 15:52
  • $\begingroup$ Thanks mate this would help $\endgroup$ – kroner Oct 4 '15 at 15:52
  • $\begingroup$ I added the new answer. Look it up. :) $\endgroup$ – H. R. Oct 4 '15 at 16:26
  • $\begingroup$ what if the last term has no $x$ in it? how will get the solution as a Bessel function? $\endgroup$ – John Oct 27 '18 at 6:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.